Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define ld long double
int main()
{
	int n;
	cin>>n;
	vector<ld>dp(n+1);
	dp[0]=1;
	for(int coins = 1;coins<=n;coins++)
	{
		ld c;
		cin>>c;
		for(int j=coins;j>=0;j--)
		{
			dp[j] = (j==0?0:dp[j-1]*c)+dp[j]*(1-c);
		}
	}
	ld ans = 0;
	for(int heads = 0;heads<=n;heads++)
	{
		int tails = n-heads;
		if(heads>tails)
		{
			ans += dp[heads];
		}
	}
	cout<<setprecision(10)<<ans<<"\n";
	return 0;
}